1
On Choosing and Using Control Volumes: Six Ways of
Applying the Integral Mass Conservation Theorem to a
Simple Problem
Ain A. Sonin, MIT
2001, 6 pages
Reference: Ain A. Sonin, Fundamental Laws of Motion for Particles, Material Volumes,
and Control Volumes, 2001
We shall use a very simple example to illustrate the variety of ways in which a control
volume theorem can be applied to a particular application, depending on the choice of
control volume and which of the two alternative forms of the control volume theorem is
used. The exercise provides a few basic insights into the thought processes that are used
in control volume analysis.
Figure 1 depicts something like a cylindrical syringe, or a grease gun, in which a
solid piston of radius R 1 is pushed at a speed U(t) into a fluid-filled cylinder with the
same internal radius, forcing the fluid out through a tube with internal radius R2 and
length L. The piston, cylinder, and tube are inflexible and made of material with density
ρs; the fluid has density ρ and can be considered incompressible.
Fig. 1: The system and the control volume (broken red line) for Methods 1 and 2. This control volume is
fixed in the inertial reference frame of the cylinder.
2
Given the aforementioned quantities, what is the average flow speed V(t) of the
fluid at the exit plane? The governing principle is clearly mass conservation, which can
be written for a control volume in two alternative forms,
Form A
Form B
d
dt
∫ ρdV + ∫ ρv
dA = 0
(1)
∂ρ
dV + ∫ ρvn dA = 0
∂t
CV (t )
CS (t )
(2)
CV( t )
rn
CS(t )
∫
r r
where vn = v ⋅ n = vcos θ is the outward normal velocity component of the fluid at the
v v v
control surface, and vrn = (v − vc ) ⋅ n is the outward normal component of the fluid
velocity relative to the control surface at that point. The answer is simple:
V(t) =
R12 U(t)
.
2
R2
(3)
Let us look at some of the different ways this result can be obtained.
In all the examples that follow, the reference frame is fixed in the solid cylinder.
Method 1
Control volume: As shown in Fig. 1, with all points of the control surface held fixed in
the chosen reference frame. Note that the top of this CS is placed just below the piston
face, that is, we choose to look at the situation at an instant before the piston’s face
reaches the fixed elevation of the top of our control surface. All parts of the CS reside in
the fluid, where the density is ρ.
Equation: We shall apply Form A of mass conservation,
d
dt
∫ ρdV + ∫ ρv
CV( t )
rn
dA = 0
(1)
CS(t )
Analysis: The first term in (1) is zero because the fluid density is invariant and our
control volume has a fixed volume, so that the mass contained within it is invariant. (We
are considering a time just before the piston’s face penetrates the top of the CS!) In the
3
second term, vn = −U (t) at the top of the CS where the piston face pushes the fluid down
into the CV through a stationary CS with area πR12 . At the exit plane the area is πR22 and
the velocity is vn = V(t ) on the average. Elsewhere on the CS, vn = 0 . Equation (1) thus
gives
ρVπR22 + ρ ( −U )πR12 = 0 ,
(4)
which is the same as (3).
Method 2
Control volume: Same as in Method 1.
Equation: Form B of the mass conservation equation,
∂ρ
dV + ∫ ρvn dA = 0 .
∂t
CV (t )
CS (t )
∫
(2)
Analysis: In this case the volume integral is zero because the density is invariant with
time. The integral over the CS is exactly the same as in Method 1, and we end up with (3)
again.
Method 3
Control volume: As shown in Fig. 2, i.e. similar to the one in Fig. 1 except that the top
surface moves downward at the piston speed U, keeping just a hair ahead of the piston
face, that is, the CV height h(t) is such that
dh
= −U(t) .
dt
(5)
Equation: Form A of the mass conservation theorem,
d
dt
∫ ρdV + ∫ ρv
CV( t )
CS(t )
rn
dA = 0
(1)
4
Analysis: In this case the first term in (1) is non-zero because the CV’s volume changes
as the piston moves down and h decreases. The outflow term in the surface integral is the
same as before, but there is no inflow into the CV from the top because the fluid moves
down at the same speed as the piston. Equation (1) yields
d
ρπR12 h(t) + ρπR22 L] + ρV(t)πR22 = 0
[
dt
(6)
Fig. 2: Control volume for Methods 3 and 4 (broken red line). The top moves down under the piston.
where the quantity in square brackets in (6) is the volume integral of the density within
the CV. We evaluate the time derivative in (6) using (5), and again obtain (3).
Method 4
Control volume: Same as in Method 3. See Fig. 2.
Equation: Form B of the mass conservation theorem,
∂ρ
dV + ∫ ρvn dA = 0 .
∂t
CV (t )
CS (t )
∫
(2)
Analysis: The first term is zero because the density does not vary with time. The surface
integral involves the normal fluid velocity, not the normal relative velocity. (The control
surface velocity appears nowhere in (2)!) Equation (2) reads
5
0 + [−ρU (t)πR1 + ρV (t)πR2 ] = 0
2
2
(7)
where the term in square brackets is the surface integral.
Method 5
Control volume: This time we deliberately pick a less thoughtful control volume—one
with a fixed upper bounding surface that cuts through the piston at a fixed elevation ho
(Fig. 3).
Equation: Form A of the mass conservation theorem,
d
dt
∫ ρdV + ∫ ρv
CV( t )
rn
dA = 0
(1)
CS(t )
Fig. 3: Control volume for Methods 5 and 6 (broken red line). The control volume is fixed.
Analysis:
d
ρπR12 h + ρsπR12 (ho − h ) + ρπR22 L] + [− ρsUπR12 + ρVπR22 ] = 0
[
dt
(8)
6
Here h(t) is the elevation of the piston face, the first term in square brackets is the volume
integral of density, and the second term in square brackets is the surface integral of mass
flux.. Evaluating the time derivative using (5), we find that (8) reduces to (3).
Method 6
Control volume: Same as in Method 5.
Equation used: Form B of the mass conservation theorem,
∂ρ
dV + ∫ ρvn dA = 0 .
∂t
CV (t )
CS (t )
∫
(2)
Analysis: When we now try to apply Form B, we encounter a difficulty: the partial time
derivative of density is zero everywhere except at the piston face, where it is singular. At
any height z inside the cylinder, the density is equal to the fluid density ρ before the
piston arrives, jumps discontinuously to the solid density ρs at the instant the piston
arrives, and stays at that value thereafter. If for example the piston velocity is constant,
this can be expressed as
ρ =ρ
z < ho − Ut
∂ρ
→∞
∂t
z = ho − Ut
ρ = ρs
z > ho − Ut
(9)
The singularity makes it difficult to evaluate the volume integral of the time derivative of
the density. It is possible to integrate this particular singularity by representing the partial
time derivative of the density with a suitable Dirac delta function. We will not do that
here, for it requires an introduction to generalized functions (see, however, Problem 3.7
for an analogous problem with a solution provided). A better way to proceed with this
CV choice (if one is fixated on it) is to use Form A of the mass conservation theorem,
that is, Method 5.
MIT OpenCourseWare
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2.25 Advanced Fluid Mechanics
Fall 2013
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